$g(n) = 2n+3(h(n))$ $f(x) = 6x-3+4(g(x))$ $h(t) = -3t$ $ g(h(1)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = (-3)(1)$ $h(1) = -3$ Now we know that $h(1) = -3$ . Let's solve for $g(h(1))$ , which is $g(-3)$ $g(-3) = (2)(-3)+3(h(-3))$ To solve for the value of $g$ , we need to solve for the value of $h(-3)$ $h(-3) = (-3)(-3)$ $h(-3) = 9$ That means $g(-3) = (2)(-3)+(3)(9)$ $g(-3) = 21$